Chapter 1: Vectors, Lines and Planes
Section 1.4: Cross Product
If A=a1 i+a2 j+a3 k and B=b1 i+b2 j+b3 k are two vectors in ℝ3, the cross product A×B is the vector detailed in Definition 1.4.1.
Because this product yields a vector, some texts call it the vector product of two vectors.
Definition 1.4.1 - Cross Product
A×B≡a2b3−a3b2i+a3b1−a1b3j+a1b2−a2b1k = a2b3−a3b2a3b1−a1b3a1b2−a2b1
Because it is no small feat to remember the order of all the subscripts in Definition 1.4.1, the following mnemonic (or memory device) is useful as the computational tool by which a cross product can be computed by hand.
A×B≡ a2b3−a3b2a3b1−a1b3a1b2−a2b1 = ijka1a2a3b1b2b3 = i a2a3b2b3−j |a1a3b1b3|+k a1a2b1b2
The vector A×B can be obtained as the determinant of the 3 × 3 matrix that has the unit basis vectors i,j,k as the entries in the first row, the components of A in the second row, and of B in the third. The determinant is evaluated by a first-row Laplace expansion in which the three resulting 2 × 2 determinants are called minors. The signs in front of the minors alternate, and a signed minor is called a cofactor. The ith minor for the elements in the first row is the determinant of the array obtained by deleting that row and the ith column.
Note: An internet search for "Laplace expansion 3x3" will bring up a number of sites that review the process of evaluating a determinant by the row-expansion process.
Table 1.4.1 lists some of the cross product's properties that can be extracted from Definition 1.4.1.
A×B = A B sin(θ)
A×B is orthogonal to both A and B
A, B, and A×B form a right-handed system of vectors:
c A×B=cA×B=A×c B
A×B+C=A×B+A×C and A+B×C=A×C+B×C
Associative Rule - 1
Associative Rule - 2
Table 1.4.1 Properties of the cross product
A right-handed system of the three vectors A, B, and A×B is modeled on the threads of a standard right-threaded screw or bolt that "advances" when rotated clockwise. (Woodworkers, plumbers, mechanics will tell their apprentices "right = tight, left = loose".) If the vector A is rotated into B, the direction of A×B would be that of the advance of the screw turned in the same direction. In other words, if A and B were viewed in a plane facing the viewer with A to the left of B, then A×B would point away from the viewer because the rotation of A into B would be clockwise; with A to the right of B, A×B would point toward the viewer.
Some texts define the cross product by the length, orthogonality, and right-handedness properties, then derive the expression in Definition 1.4.1.
For the vectors A=3 i−2 j+4 k and B=5 i+7 j−6 k, and the scalar c,
Obtain θ, the angle between A and B.
Verify computationally that A×A=0.
Verify computationally that A×B=A B sin(θ).
Verify computationally that B×A= −A×B.
Verify that both A and B are perpendicular to A×B.
Verify that A·B2+A×B2 = A B2.
Verify that A·B2−A×B2 = A B2cos(2 θ).
Show that A×A×B≠A×A×B=0.
Verify that c A×B=cA×B=A×c B.
Verify that A−B×A+B=2A×B.
For the vectors A=3 i−2 j+4 k, B=5 i+7 j−6 k, and C=4 i−3 j−5 k,
Verify the distributive property A×B+C=A×B+A×C.
Verify the distributive property A+B×C=A×C+B×C.
Show that A×B×C≠A×B×C.
Verify the identity A×B×C+B×C×A+C×A×B=0.
Verify the identity A×B×C=A·CB−A·BC.
Verify the identity A×B×C=A·CB−B·CA.
For the vectors A, B, and C of Example 1.4.2, and V=2 i−5 j+3 k, verify the identity A×B·C×V=A·CA·VB·CB·V.
Find a unit vector orthogonal to the plane containing the vectors A=5 i−3 j+7 k and B=6 i+2 j−3 k.
Find a unit vector orthogonal to the plane containing the points P:1,2,−1, Q:3,1,2, R:4,−5,1.
Taking A=a1 i+a2 j+a3 k and B=b1 i+b2 j+b3 k, and using Definition 1.4.1, verify that A×B is orthogonal to both A and B.
Taking A=a1 i+a2 j+a3 k, B=b1 i+b2 j+b3 k, and C=c1 i+c2 j+c3 k, and using Definition 1.4.1, verify the two distributive laws in Table 1.4.1.
Starting with the length, orthogonality, and right-handedness properties in Table 1.4.1, derive the expression for the cross product stated in Definition 1.4.1.
Starting with Definition 1.4.1, derive the length property, A×B = A B sin(θ).
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